Answer
$\lim_{\theta\to0}(2-\cot\theta)$ does not exist.
Work Step by Step
$$A=\lim_{\theta\to0}(2-\cot\theta)=2-\lim_{\theta\to0}\cot\theta=2-\lim_{\theta\to0}\frac{\cos\theta}{\sin\theta}$$
As $\theta\to0^+$, $\cos\theta$ approaches $\cos0=1\gt0$, while $\sin\theta$ approaches $\sin\theta=0$ from the right, where $\sin\theta\gt0$.
As $\theta\to0^-$, $\cos\theta$ approaches $\cos0=1\gt0$, while $\sin\theta$ approaches $\sin\theta=0$ from the left, where $\sin\theta\lt0$.
Therefore, as $\theta\to0^+$, $\cos\theta/\sin\theta$ will approach $\infty$; while as $\theta\to0^-$, $\cos\theta/\sin\theta$ will approach $-\infty$. In other words,
$$\lim_{\theta\to0^+}\frac{\cos\theta}{\sin\theta}=\infty\hspace{1cm}\lim_{\theta\to0^-}\frac{\cos\theta}{\sin\theta}=-\infty$$
Since $\lim_{\theta\to0^+}\frac{\cos\theta}{\sin\theta}\ne\lim_{\theta\to0^-}\frac{\cos\theta}{\sin\theta}$, $\lim_{\theta\to0}\frac{\cos\theta}{\sin\theta}$ does not exist. So $\lim_{\theta\to0}(2-\cot\theta)$ does not exist.
